N-Epsilon Proof---HELP

Caller

Poland8075 Posts

October 19 2008 20:00 GMT

Hi everybody

I have a Calculus midterm tomorrow and I have no idea what to do for an n-epsilon proof. Delta epsilon isn't so bad right now, but n-epsilon is.

From what I understand, according the the precise definition of a limit,

lim x->infinity of a sequence a_n --> L

and I have to set it so absolute value |a_n-L| < epsilon
or something like that.

And I have no idea what the hell this means.

Halp anybody???

Kwidowmaker

Canada978 Posts

October 19 2008 20:06 GMT

You have to show that as lim n -> infinity, the absolute difference between a_n and L becomes smaller than any positive number epsilon. Basically, you have to show that a_n approaches L in the same sense that lim x -> 5 (x + 2) = 7.

Muirhead

United States557 Posts

October 19 2008 20:12 GMT

lim_(n---> infinity) a_n=L
if and only if
For any positive real number epsilon, there exists a positive integer N such that |a_m-L|<epsilon whenever m>N.

Caller

Poland8075 Posts

October 19 2008 20:24 GMT

So how would I go about doing the actual proof itself? I never learned how to go about a calculus proof, can anybody give me an example?

zeks

Canada1068 Posts

October 19 2008 21:03 GMT

write it out 10 times and then just regurgitate it during the midterm =/

i agree with muirhead's proof, but you might want to check your notes/textbook because you might be used to another set of notation

micronesia

United States24805 Posts

October 19 2008 21:09 GMT

On October 20 2008 06:03 zeks wrote:
write it out 10 times and then just regurgitate it during the midterm =/

i agree with muirhead's proof, but you might want to check your notes/textbook because you might be used to another set of notation

A good professor will prevent you from succeeding on a question about this simply by regurgitating the exact proof. One possible way is your second statement. Also, theoretical questions that require a moderate understanding of the principles behind the proof are another way the professor can force students to truly learn it.

Kwidowmaker

Canada978 Posts

October 19 2008 21:23 GMT

you can only prove it for converging sequences, a_n.

Provide a converging sequence, a_n, and we can show a proof for an example. A monotone decreasing sequence is easiest.

fight_or_flight

United States3988 Posts

October 19 2008 21:53 GMT

On October 20 2008 05:00 Caller wrote:
Hi everybody

I have a Calculus midterm tomorrow and I have no idea what to do for an n-epsilon proof. Delta epsilon isn't so bad right now, but n-epsilon is.

From what I understand, according the the precise definition of a limit,

lim x->infinity of a sequence a_n --> L

and I have to set it so absolute value |a_n-L| < epsilon
or something like that.

And I have no idea what the hell this means.

Halp anybody???

Since epsilon is a constant number, and |a_n-L| should always be getting smaller (because its converging), you should always be able to reach a point where |a_n-L| < epsilon for some (probably large) n.

Wonders

Australia753 Posts

October 19 2008 22:43 GMT

Say a sequence converges to a particular limit. What this means intuitively is that EVENTUALLY, the terms get as close as you want to that limit, if you go far enough along the sequence.

So to formalize this, you let epsilon be any number, which is as small as you want it to be (but not zero). Then if you go far enough along the sequence, then the difference between terms in the sequence and the limit are going to be smaller than epsilon (which is as small as you like). That is, there exists some possibly huge number M, such that all terms in the sequence after M are going to be closer to the limit than epsilon, or |a_n - L| < epsilon for all n>M.

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